Problem: Let $h(x)=\dfrac{\ln(x)}{x^2}$. $h'(x)=$
Explanation: $h(x)$ is the quotient of two, more basic, expressions: $\ln(x)$ and $x^2$. Therefore, the derivative of $h$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{\ln(x)}{x^2}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\ln(x))x^2-\ln(x)\dfrac{d}{dx}(x^2)}{(x^2)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\dfrac1x\cdot x^2-\ln(x)\cdot 2x}{(x^2)^2}&&\gray{\text{Differentiate }\ln(x)\text{ and }x^2} \\\\ &=\dfrac{x-2x\ln(x)}{x^4}&&\gray{\text{Simplify}} \\\\ &=\dfrac{x(1-2\ln(x))}{x^4}&&\gray{\text{Factor out }x} \\\\ &=\dfrac{1-2\ln(x)}{x^3}&&\gray{\text{Cancel common factors}} \end{aligned}$ In conclusion, $h'(x)=\dfrac{1-2\ln(x)}{x^3}$ or any other equivalent form.